Prasanna

Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Polynomials with Answers Solutions

Extra Questions for Class 9 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 9 Extra Questions Very Short Answer Type

Polynomials Class 9 Extra Questions Question 1.
Factorise : 125x³ – 64y³
Solution:
125x³– 6443 = (5x)³ – (4y)³
By using a3 – b3 = (a – b) (a² + ab + b²), we obtain
125x³– 64y³ = (5x – 4y) (25x² + 20xy + 16y²)

Class 9 Polynomials Extra Questions Question 2.
Find the value of (x + y)² + (x – y)².
Solution:
(x + y)² + (x – y)² = x² + y² + 2xy + x² + y² – 2xy
= 2x² + 2y² = 21x² + y²)

Square polynomial root calculator. grade 10 math linear algebra substitution word problems.

Polynomial Class 9 Extra Questions Question 3.
If p(x)= x² – 2√2x+1, then find the value of p(2√2)
Solution:
Put x = 2√2 in p(x), we obtain
p(2√2) = (2√2)² – 2√2(2√2) + 1 = (2√2)² – (2√2)² + 1 = 1

Polynomials Extra Questions Class 9 Question 4.
Find the value of m, if x + 4 is a factor of the polynomial x² + 3x + m.
Solution:
Let p(x) = x² + 3x + m
Since (x + 4) or (x – (-4)} is a factor of p(x).
∴ p(-4) = 0
⇒ (-4)² + 3(-4) + m = 0
⇒ 16 – 12 + m = 0
⇒ m = -4

Class 9 Maths Chapter 2 Extra Questions Question 5.
Find the remainder when x³+ x² + x + 1 is divided by x – \(\frac{1}{2}\) using remainder theorem.
Solution:
Let p(x) = x³+ x² + x + 1 and q(x) = x – \(\frac{1}{2}\)
Here, p(x) is divided by q(x)
∴ By using remainder theorem, we have

Extra Questions For Class 9 Maths Chapter 2 Polynomials With Answers Question 6.
Find the common factor in the quadratic polynomials x² + 8x + 15 and x² + 3x – 10.
Solution:
x² + 8x + 15 = x² + 5x + 3x + 15 = (x + 3) (x + 5)
x² + 3x – 10 = x² + 5x – 2x – 10 = (x – 2) (x + 5)
Clearly, the common factor is x + 5.

Polynomials Class 9 Extra Questions Short Answer Type 1

Extra Questions On Polynomials Class 9 Question 1.
Expand :
(i) (y – √3)²
(ii) (x – 2y – 3z)²
Solution: (i)
(y – √3)² = y² -2 × y × √3 + (√3)² = y² – 2√3 y + 3 (x – 2y – 3z)²
= x² + 1 – 2y)² + (-3z)² + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x
= x² + 4y² + 9z² – 4xy + 12yz – 6zx

Extra Questions Of Polynomials Class 9 Question 2.
If x + = \(\frac{1}{x}\) = 7, then find the value of x³ + \(\frac{1}{x^{3}}\)
Solution:
We have x + \(\frac{1}{x}\) = 7
Cubing both sides, we have

Questions On Polynomials Class 9 Question 3.
Show that p – 1 is a factor of p¹⁰ + p⁸ + p⁶ – p⁴ – p² – 1.
Solution:
Let f(p) = p¹⁰ + p⁸ + p⁶ – p⁴ – p² – 1
Put p = 1, we obtain
f(1) = 1¹⁰ + 1⁸ + 1⁶ – 1⁴ – 1² – 1
= 1 + 1 + 1 – 1 – 1 – 1 = 0
Hence, p – 1 is a factor of p¹⁰ + p⁸ + p⁶ – p⁴ – p² – 1.

Class 9 Maths Polynomials Extra Questions Question 4.
If 3x + 2y = 12 and xy = 6, find the value of 27x³ + 8y³
Solution:
We have 3x + 2y = 12
On cubing both sides, we have
⇒ (3x + 2y)³ = 12³
⇒ (3x)³ +(2y)³ + 3 × 3x × 2y(3x + 2y) = √728
⇒ 27x³+ 8y³ + 18xy(3x + 2y) = √728
⇒ 27x³+ 8y³ + 18 × 6 × 12 = √728
⇒ 27x³+ 8y³ + 1296 = √728
⇒ 27x³+ 8y³ = √728 – 1296
⇒ 27x³+ 8y³ = 432

Polynomials Class 9 Questions With Solutions Question 5.
Factorise : 4x² + 9y² + 16z²2 + 12xy – 24 yz – 16xz.
Solution:
4x² + 9y² + 16z²2 + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)
By using a² + b² + 2ab + 2bc + 2ca = (a + b + c)², we obtain
= (2x + 3y – 4z)² = (2x + 3y – 4z) (2x + 3y – 4z)

Class 9 Maths Ch 2 Extra Questions Question 6.
Factorise : 1 – 2ab – (a² + b²).
Solution:
1 – 2ab – (a² + b²) = 1 – (a² + b² + 2ab)
= 1² – (a + b)²
= (1 + a + b) (1 – a – b) [∵ x² – y² = (x + y)(x – y)]

Polynomials Class 9 Extra Questions Short Answer Type 2

Polynomials Class 9 Extra Questions With Answers Pdf Question 1.
Factorise :

Solution:

Important Questions Of Polynomials For Class 9 Question 2.
Factorise 64a³ – 27b³ – 144a²b + 108ab².
Solution:
64a² – 27b² – 144a²b + 108ab²
= (4a)³ – (3b)³ – 36ab(4a – 3b)
= (40)² – (3b)³ – 3 × 4a × 3b (4a – 3b)
= (4a – 3b)³ [∵ (x – y)³ = x³ – y³ – 3xy(x – y)]
= (40 – 3b) (4a – 3b) (4a – 3b)

Polynomial Extra Questions Class 9 Question 3.
What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
Volume = 12ky² + 8ky – 20k.
Solution:
We have, volume = 12ky² + 8ky – 20k
= 4k(3y² + 2y – 5) = 4k(3y² + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1)
∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.

Polynomials Class 9 Extra Questions With Solutions Question 4.
If p(x) = x³ + 3x² – 2x + 4, then find the value of p(2) + p(-2) – P(0).
Solution:
Here, p(x) = x³+ 3x² – 2x + 4
Now, p(2) = 2³ + 3(2)² – 2(2) + 4
= 8 + 12 – 4 + 4 = 20
p(-2) = (-2)³ + 3(-2)² – 21 – 2) + 4
= 8 + 12 + 4 + 4 = 12
and p(0) = 0 + 0 – 0 + 4 = 4
∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.

Class 9 Chapter 2 Maths Extra Questions Question 5.
If one zero of the polynomial x² – √3x + 40 is 5, which is the other zero ?
Solution:
Let p(x) = x² – √3x + 40
= x² – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
Now, for zeroes of given polynomial, put p(x) = 0
∴ (x – 5) (x – 8) = 0
⇒ x = 5 or x = 8
Hence, other zero is 8.

Polynomials Class 9 Important Questions Question 6.
Simplify:

Solution:

Extra Questions For Class 9 Maths Polynomials Question 7.
If one zero of the polynomial x² – √3x + 40 is 5, which is the other zero ?
Solution:
Let
p(x) = x² – √3x + 40
= x² – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
Now, for zeroes of given polynomial, put p(x) = 0
∴ (x – 5) (x – 8) = 0
x = 5 or x = 8
⇒ Hence, other zero is 8.

Polynomials Class 9 Extra Questions Long Answer Type

Ch 2 Maths Class 9 Extra Questions Question 1.
Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b + c) (c + a).
Solution:
L.H.S. = (a + b + c)³ – a³ – b³ – c³
= {(a + b + c)³ – 3} – {b³ + c³}
= (a + b + c – a) {(a + b + c)² + a² + a(a + b + c)} – (b + c) (b² + c² – bc)
= (b + c) {a² + b² + 2 + 2ab + 2bc + 2ca + a² + a² + ab + ac – b² – a² + bc)
= (b + c) (3a² + 3ab + 3bc + 3ca}
= 3(b + c) {a² + ab + bc + ca}
= 31b + c) {{a² + ca) + (ab + bc)}
= 3(b + c) {a(a + c) + b(a + c)}
= 3(b + c)(a + c) (a + b)
= 3(a + b)(b + c) (c + a) = R.H.S.

Class 9 Maths Polynomials Extra Questions With Solutions Question 2.
Factorise : (m + 2n)² x² – 22x (m + 2n) + 72.
Solution:
Let m + 2n = a
∴ (m + 2n)² x² – 22x (m + 2n) + 72 = a²x² – 22ax + 72
= a²x² – 18ax – 4ax + 72
= ax(ax – 18) – 4(ax – 18)
= (ax – 4) (ax – 18)
= {(m + 2n)x – 4)} {(m + 2n)x – 18)}
= (mx + 2nx – 4) (mx + 2nx – 18).

Question 3.
If x – 3 is a factor of x² – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k.
Solution:
Here, x – 3 is a factor of x² – kx + 12
∴ By factor theorem, putting x = 3, we have remainder 0.
⇒ (3)² – k(3) + 12 = 0
⇒ 9 – 3k + 12 = 0
⇒ 3k = 21
⇒ k = 7
Now, x² – 7x + 12 = x² – 3x – 4x + 12
= x(x – 3) – 4(x – 3)
= (x – 3) (x – 4)
Hence, the value of k is 7 and other factor is x – 4.

Question 4.
Find a and b so that the polynomial x³– 10x² + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).
Solution:
Let p(x) = x³– 10x² + ax + b
Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2).
∴ By putting x = 1, we obtain
(1)³ – 10(1)² + a(1) + b = 0
⇒ a + b = 9
And by putting x = 2, we obtain
(2)³ – 10(2)² + a(2) + b = 0
8 – 40 + 2a + b = 0
⇒ 2a + b = 32
Subtracting (i) from (ii), we have
a = 23
From (i), we have 23 + b = 9 = b = -14
Hence, the values of a and b are a = 23 and b = -14

Question 5.
Factorise : x² – 6x² + 11x – 6.
Solution:
Let p(x) = x² – 6x² + 11x – 6
Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6
By putting x = 1, we have
p(1) = (1)³ – 6(1)² + 11(1) – 6 = 1 – 6 + 11 -6 = 0
∴ (x – 1) is a factor of p(x)
By putting x = 2, we have
p(2) = (2)³ – 6(2)² + 11(2) – 6 = 8 – 24 + 22 – 6 = 0
∴ (x – 2) is a factor of p(x)
By putting x = 3, we have
p(3) = (3)³ – 6(3)² + 11(3) – 6 = 27 – 54 + 33 – 6 = 0
∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors.
∴ x³ – 6x² + 11x – 6 = k (x – 1) (x – 2) (x – 3)
By putting x = 0, we obtain
0 – 0 + 0 – 6 = k (-1) (-2) (3)
-6 = -6k
k = 1
Hence, x³ – 6x² + 11x – 6 = (x – 1) (x – 2)(x – 3).

Question 6.
Show that \(\frac{1}{3}\)and \(\frac{4}{3}\) are zeroes of the polynomial 9x³ – 6x² – 11x + 4. Also, find the third zero of the polynomial.
Solution:
Let p(x) = 9x³– 6x² – 11x + 4
Put x = \(\frac{1}{3}\), we have

Thus, x = \(\frac{4}{3}\) is another zero of the polynomial p(x). Since x = \(\frac{1}{3}\) and x = \(\frac{4}{3}\) are the zeroes of p(x), therefore, \(\left(x-\frac{1}{3}\right)\) \(\left(x-\frac{4}{3}\right)\) (3x – 1) (3x – 4) or 9x² – 15x + 4 exactly divides p(x).
⇒ 9x³ – 6x² – 11x + 4 = (9x² – 15x + 4) (x + 1)
Hence, x = -1 is its third zero.

Question 7.
Factorise : 6x² – 5x² – √3x + 12
Solution:
Let p(x) = 6x³– 5x² – √3x + 12
Here, constant term of p(x) is 12 and factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12.
By putting x = 1, we have
p(1) = 6(1)³ – 5(1)² – √3(1) + 12 = 6 – 5 – √3 + 12 = 0
∴ (x – 1) is a factor of p(x).
Now, by long division, we have

Thus,
p(x) = (x – 1) (6x² + x – 12)
p(x) = (x – 1) (6x² + 9x – 8x – 12)
p(x) = (x – 1) {3x (2x + 3) – 4(2x + 3)}
p(x) = (x – 1) (3x – 4) (2x + 3).

Polynomials Class 9 Extra Questions HOTS

Question 1.
What must be added to polynomial f(x) = x⁴ + 2x² – 2x² + x – 1 so that resulting polynomial is exactly divisible by x² + 2x – 3?
Solution:

Here, remainder = -x + 2
To make remainder = 0, we must add -(remainder) in the polynomial
i.e., -(-x + 2) i.e., x – 2
Hence, x4 + 2x³ – 2x² + x – 1 + (x – 2)
Here, polynomial = x⁴ + 2x³– 2x² + 2x – 3 and required addition is (x – 2).

Question 2.
If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?.
Solution:
Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0
x³+ √3 + x³= 3(x)(y)(z)
= 3(2 – √3)(√3 – √7)(√7 – 2)
= 3(2√3 – 2√7 – 3 + √21)(√7 – 2)
= 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21)
= 3(3√3 + √7 – 8)

Question 3.
If (x – a) is a factor of the polynomials x² + px – q and x² + rx – t, then prove that a = \(\frac{t-q}{r-p}\)
Solution:
Let f(x) = x + px -q and g(x) = x² + x – t
Since x-a is factor of both f(x) and g(x)
⇒ f(a) = g(a) = 0
Now, here f(a) = a² + pa – q and
g(a) = a² + ra- t
⇒ a² + pa – q = a + ra – t (considering f(a) = g(a)]
⇒ pa – q = ra – t
⇒ ra – pa = t – q
⇒ a(r – p) = t – q
a = \(\frac{t-q}{r-p}\)

Polynomials Class 9 Extra Questions Value Based (VBQs)

Question 1.
If a teacher divides a material of volume 27x³ + 54x² + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ?
Solution:
We know that, √olume = Length × Breadth × Height
Now, 27x³+ 54x² + 36x + 8
= (3x)³ + 3(3x)²(2) + 3(3x)(2)² + (2)³
= (3x + 2)² = (3x + 2) (3x + 2) (3x + 2)
Thus, volume = (3x + 2) (3x + 2) (3x + 2)
Yes, it is possible to find the quantity of material. (3x + 2) units.
Cube.
Apply knowledge and use of example for clarity of subject, student friendly.

Question 2.
In a camp organised by the students of class-9 to donate amount collected to flood victims of Kashmir. At the time of payment of a juice glass at one stall of juice, stall holder asked the students to pay the remainder of x³+ 3x² + 3x + 1 divided by \(\left(x-\frac{1}{2}\right)\) What is the price of the juice at the stall ? Which value is depicted by class-9 students by organising such camps ?
Solution:
Let
p(x) = x² + 3x² + 3x + 1 and

By long division method, we have

Remainder = \(\frac{27}{8}\) or 3 \(\frac{3}{8}\)
Thus, price of the juice glass is ₹ 3 \(\frac{3}{8}\)
Caring, kindness, social welfare and helping in development of the needy.

Here we are providing Online Education for Vocation Extra Questions and Answers Class 6 English Honeysuckle, Extra Questions for Class 6 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-6-english/

Online Education for Vocation Extra Questions and Answers Class 6 English Honeysuckle

Vocation Extra Questions and Answers Short Answer Type

Your partner and you may now be able to answer these questions.

Vocation Poem Extra Questions And Answers Class 6 Question 1.
Who is the speaker in the poem? Who are the people the speaker meets? What are they doing?
Answer:
A school-going child speaks in the poem. The speaker meets a hawker, a gardener and a watchman. The hawker sells bangles. The gardener digs the ground. The watchman keeps a watch in the street at night.

Vocation Class 6 Questions And Answers Question 2.
What wishes does the child in the poem make? Why does the child want to be a hawker, a gardener, or a watchman? Pick out the lines in each stanza, which tell us this.
Answer:
The child in the poem wants to be a hawker, a gardener, and a watchman. When he looks at the hawker, he wishes he could also spend his day on the road crying “Bangles, crystal bangles!” He feels that there is nothing to hurry the hawker on. There is no road he must take, no place he must go to, and no fixed time when he must come home. These are the things that he cannot do himself and therefore, he wants to be a hawker so that he could do all these things.

Next, he wishes he was a gardener because a gardener does what he likes with his spade. He soils his clothes with dust. Nobody scolds him if he gets baked in the sun or gets wet. Therefore, if the little child was a gardener, nobody would stop him from digging. Finally, he sees the watchman and wants to be like the watchman so that he could walk through dark and lonely streets all night with his lantern and chase shadows.

When he is put to bed and is not allowed to roam outside, he see the watchman swinging his lantern with his shadow at his side and he feels that the watchman never even once has to go to bed in his entire life. Therefore, he wants to be a hawker, a gardener, and a watchman so that he could do all the things they did as he could not do them being a child.

Vocation Poem Class 6 Questions And Answers Pdf Question 3.
From the way the child envies the hawker, the gardener and the watchman, we can guess that there are many things the child has to do, or must not do. Make a list of the do’s and don’ts that the child doesn’t like. The first line is done for you.

The child must
come home at a fixed time.
………………..
……………….

The child must not
get his clothes dirty in the dust.
……………..
……………….
Now add to the list your own complaints about the things you have to do, or must not do.
Answer:
The child must
come home at a fixed time
Go to bed when it get dark
Go to school everyday
Please do yourself.

The child must not
get his clothes dirty in the dust.
Get baked in the sun or get wet
Walk on dark and lonely street at night.

Vocation Question Answer Class 6 Question 4.
Like the child in the poem, you perhaps have your own wishes for yourself. Talk to your friend, using “I wish I were…”
Answer:
Please do yourself.

Vocation Chapter Question Answer Class 6 Question 5.
Find out the different kinds of work done by the people in your neighbourhood. Make different cards for different kinds of work. You can make the card colourful with pictures of the persons doing the work.
Answer:
Please do yourself.

Vocation Extra Questions and Answers Reference of Context

Vocation Poem Class 6 Questions And Answers Question 1.
When the gong sounds ten in the morning and
I walk to school by our lane,
Every day I meet the hawker crying, “Bangles,
crystal bangles!”
There is nothing to hurry him on, there is no
road he must take, no place he must go to, no
time when he must come home.
I wish I were a hawker, spending my day in
the road, crying, “Bangles, crystal bangles!”

Explanation
The poet walks to the school lane when the gong rang at ten in the morning. On the way, he meets a Bangle seller who sells crystal bangles. He need not to hurry and worry as the poet has to. He is jealous of the hawker because he is not restricted from going anywhere. He is also not forced to reach home on time. The poet wishes to be a hawker and sell bangles in the streets.

(i) Who wrote the poem?
(ii) At what time does the poet go to school?
(iii) What does the hawker cry to sell bangles?
(iv) What does the poet say about the life of the hawker?
(v) Give the meaning of‘crystal’.
Answer:
(i) Rabindranath Tagore wrote the poem.
(ii) The poet goes to school at 10 a.m.
(iii) The hawker cries “Bangles, crystal bangles” to sell his bangles.
(iv) The poet says that the hawker has freedom.
(v) Highly transparent glass.

Vocation Poem Questions And Answers Class 6 Question 2.
When at four in the afternoon I come back from the school,
I can see through the gate of that house the gardener digging the ground.
He does what he likes with his spade, he soils
his clothes with dust, nobody takes him to
task, if he gets baked in the sun or gets wet.
I wish I were a gardener digging away at the
garden with nobody to stop me from digging.

Explanation
The poet witnesses the freedom of choice of the gardener whom he comes across on the way home. The gardener digs the ground with the spade. His dress is soiled with dust, but still he is not scolded by anyone. He is free to be in sunlight or get wet in the rains. The poet wishes to bask in the sun and get drenched in the rains. He wants to enjoy freedom of digging the ground.

(i) Whom does the poet see at four in the afternoon?
(ii) Where does the poet see him?
(iii) What does the gardener do with the spade?
(iv) What is the wish of the poet?
(v) Give the meaning of baked.
Answer:
(i) The poet sees the gardener at four in the afternoon.
(ii) The poet sees him through the gate of a house.
(iii) The gardener does what he likes with his spade.
(iv) The poet wishes that nobody should stop him if he were a gardener.
(v) (a) Food cooked by dry heat in an oven.
(b) Stay in sunlight for long hours.

Vocation Poem Class 6 Extra Questions And Answers Question 3.
Just as it gets dark in the evening and my
mother sends me to bed,
I can see through my open window the
watchman walking up and down.
The lane is dark and lonely, and the street lamp
stands like a giant with one red give in its head.
The watchman swings his lantern and walks
with his shadow at his side, and never once
goes to bed in his life.
I wish I were a watchman walking the street
all night, chasing the shadow with my lantern.

Explanation
The poet’s mother compels the poet to slip into bed. The poet sees through his window a watchman doing his duty. The lanes are dark and desolate. The poet describes that he looks like a giant with one red eye in its head. The lantern saving here and there in his hand. The poet wants to chase his shadow like him.

(i) Who sends the poet to bed?
(ii) Whom does he see through his window?
(iii) How does the poet describe the street lamp?
(iv) What was his reason of comparison?
(v) Choose the word similar to ‘torch’.
Answer:
(i) The poet’s mother send him to bed.
(ii) He sees a watchman through his window.
(iii) The poet describes the street lamp as a giant with one red eye in its head.
(iv) He does not wish to go to bed early and wants to roam around in streets chasing his own shadow.
(v) Lantern.

Here we are providing If I Were You Extra Questions and Answers Class 9 English Beehive, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

If I Were You Extra Questions and Answers Class 9 English Beehive

If I Were You Extra Questions and Answers Short Answer Type

If I Were You Extra Questions Question 1.
Briefly describe Gerrard’s appearance.
Answer:
Gerrard is a man of medium height and wears horn-rimmed glasses. When the play opens, he is dressed in a lounge suit and a great coat. He talks in a cultured voice and his demeanour is confident.

If I Were You Class 9 Extra Questions Question 2.
Who was the Intruder in Gerrard’s house? Why did he break into his house?
Answer:
The Intruder, who broke into Gerrard’s house, was a criminal. He had murdered a cop and was being chased by the police. He broke into Gerrard’s house with the intention of murdering him and taking on his identity to evade the police.

If I Were You Class 9 Extra Questions And Answers Question 3.
How did Gerrard behave on seeing a gun-toting stranger in his cottage?
Answer:
Gerrard kept his cool and remained absolutely unruffled when he saw the gun-totting stranger in his cottage. There was neither any panic nor any ring of tension in his voice. He remained his normal self and talked to him casually.

If I Were You Extra Question Answer Question 4.
Why does the Intruder intend to kill Gerrard?
Answer:
The Intruder is a criminal who is being chased by the police for having murdered a cop. As per his plan, the Intruder intends to kill Gerrard in order to take on his identity and escape capture by the police. In this way, he can lead a peaceful life without living in constant fear of arrest and punishment.

Extra Questions Of If I Were You Question 5.
Why does the Intruder not kill Gerrard immediately?
Answer:
The Intruder does not kill Gerrard immediately because he first wants to get all the necessary information from him. Without this information, his plan to take on Gerrard’s identity will not succeed.

Class 9 If I Were You Extra Questions Question 6.
What impression do you form of the Intruder as he comes in? Give examples to illustrate.
Answer:
The Intruder is similar in build to Gerrard enters from the right silently – revolver in hand. He is flashily dressed in an overcoat and a soft hat. He seems to be a dangerous person as he is carrying a pistol and threatens Gerrard. He claims to have killed a cop. He is mean, heartless and crafty, for he plans to kill Gerrard and assume his identity in order to escape the police. He is over-confident because he claims that Gerrard is no match for him.

If I Were You Important Questions Question 7.
“You’ll soon stop being smart.” Why did Intruder think that Gerrard was being smart?
Answer:
The Intruder thought that Gerrard was being ‘smart’ or clever and facetious because he did not show any fear at the sight of an armed man enter his house and threaten him. To the contrary, he was giving the Intruder smart answers to his questions.

Extra Questions On If I Were You Question 8.
How did the Intruder threaten Gerrard?
Answer:
The Intruder threatened Gerrard by saying that he would soon stop being smart. He would make Gerrard crawl.

Class 9 English If I Were You Extra Questions Question 9.
“I want to know a few things, see.” What sort of information did the Intruder want from Gerrard?
Answer:
The Intruder wanted personal details from Gerrard like whether he lived alone, what his Christian name was, whether he had a car and whether people visited him. All this information was necessary for the execution of his plan to dodge the police by killing Gerrard and taking on his identity.

If I Were You Extra Questions And Answers Question 10.
Why did the Intruder say, “They can’t hang me twice?”
Answer:
The Intruder said this because he was already wanted for having murdered a cop. If he managed to kill Gerrard, as per his plan, and was later arrested, it would not matter as the punishment for this murder, too, would be a hanging. They could not hang him for the two murders twice.

Class 9 English Beehive Chapter 11 Extra Questions Question 11.
How did Gerrard convince the Intruder that he was also wanted by the police?
Answer:
When Gerrard came to know about the Intruder’s plan, he kept his cool. He cooked a story to outwit him. He said he is also a murderer who was in hiding and that is why he was a mystery man who never met anyone including the tradespeople. So, if the Intruder took on his identify, he would not gain anything. He would anyway be accused of murder.

Extra Questions Of Chapter If I Were You Question 12.
Why has the Intruder chosen Gerrard as the man whose identify he wants to take?
Answer:
Gerrard is of the same height and build as the Intruder. There is some similarity in their Appearance. Moreover, Gerard was something of a mystery man who lived alone in the house and very few people visited him. He phoned in his orders, did not meet any tradespeople, and had irregular hours and habits, going away suddenly and coming back just the same. So, the Intruder thought if he killed Gerrard and took on his identity, he would not get caught. He would gain his freedom and would be free to go places. He could live without the fear of cops.

Extra Questions From If I Were You Question 13.
What did Gerrard tell the Intruder about his childhood and his present life? Was he telling the truth? Why/Why not?
Answer:
Gerrard told the Intruder that as a child, he was stolen by the gypsies and now in his thirties he was all alone in life. He was not telling the truth; he was just being funny as he wished to make it clear that he was not afraid of a gun-totting criminal. In fact, Gerrard had already started concocting stories about himself.

If I Were You Extra Question Answers Question 14.
What made Gerrard ask the Intruder, “Are you an American”?
Answer:
Gerrard asked the Intruder if he were an American as he had told him to ‘Put those paws up! ’ and had called him a ‘wise guy’. ‘Paws’ for hands and ‘guy’ for a man are colloquial American expressions. Hence, the usage of these words by the Intruder made Gerrard ask him this question.

Question 15.
The Intruder announced, “I’m going to kill you”. Was Gerrard nervous? How would you describe Gerrard’s reactions?
Answer:
Confident of his presence of mind and his ability to keep his cool in a difficult situation, Gerrard remained unruffled on being threatened by the Intruder. He remained so calm and nonchalant that the Intruder was irritated. His sense of humour also enraged the Intruder. Thus, Gerrard reacted in a calm and composed way.

Question 16.
Where did Gerrard live? Why was it a suitable place for the Intruder’s plan?
Answer:
Gerrard lived in a lonely cottage in a secluded place in the wilds of Essex. With hardly any population around, it was easy for one to commit a crime without getting detected. In addition, the place was visited by only a few people. Therefore, it was suitable for the Intruder to carry out his plan successfully over here.

Question 17.
The Intruder calls himself ‘a poor hunted rat’. Why does he do so?
Answer:
The Intruder describes himself as ‘a poor hunted rat’ because he is being chased by the police for having killed a cop and he has to keep dodging them to escape punishment. He feels he is like a rat being chased by a cat.

Question 18.
Why has the criminal been called an Intruder all through the play
Answer:
An Intruder is a who enters a place without permission in order to commit a crime. The Intruder is called an Intruder throughout the play as he has forced his way into Gerrard’s cottage, with criminal intent. He has come to murder Gerrard and to steal his identity.

Question 19.
Bring out the contrast between the Intruder and Gerrard.
Answer:
The Intruder is flashy, coarse, crude, boastful and an irritable egoist. He is overconfident and thinks that he is very smart. On the other hand, Gerrard is pleasant, cool-headed, cultured, witty and very intelligent. Despite . all this he is a modest and humble person. Although Gerrard does not brag, he proves to be much smarter and more intelligent than the Intruder and is able to outwit him.

Question 20.
Why did very few people come to Gerrard’s house? Who were the few people who visited him?
Answer:
A playwright, Gerrard needed his solitude and congenial surroundings. So, he lived all alone in a secluded place. His theatrical performances made his schedule irregular and he would go away suddenly and come back just the same. Moreover, very few people came to his house. He was visited only by his regular suppliers like the baker, the greengrocer and the milkman, but he did not meet them.

Question 21.
Gerrard tells the Intruder “A mystery I propose to explain.” What is the mystery that he proposes to explain?
Answer:
The Intruder has just told Gerrard that the people in Aylesbury refer to him as a “mystery man”. Gerrard proposes to explain the mystery about his mysterious life, his sudden comings and goings, his irregular routine, and his refusal to see the tradesmen. Gerrard has already concocted a story attributing his strange behaviour to his being a criminal wanted in many, cases of crime.

Question 22.
“This is your big surprise”. Who says these words in the play? What and where? What is the surprise?
Answer:
Gerrard says these words when the Intruder asks him to clarify how he could still be hanged after assuming Vincent Charles Gerrard’s identity. This is a surprise for the Intruder who never suspected Gerrard of being a criminal. According to his information, Gerrard seemed to be the perfect person who could be easily eliminated and then impersonated. However, Gerrard has just turned the tables on him by claiming to be a criminal on the run.

Question 23.
Why and how did Gerrard persuade the Intruder to get into the cupboard?
Answer:
Gerrard concocted a story about his own criminal background. He gave the Intruder the impression that the police were looking for him and he expected a telephone call from a friend, posted as a lookout, informing him of the police’s arrival. So when the telephone rang, he hurried the Intruder into the cupboard and told him that it was connected to the garage which was an escape route.

Question 24.
How does Gerrard propose to use the Intruder’s episode?
Answer:
Being a playwright, Gerrard is amused at being able to turn the tables on the Intruder. He finds the episode of outwitting a criminal by a clever but an innocent man amusing enough to use it as a plot for his next play.

Question 25.
Gerrard describes this encounter with the Intruder as ‘an amusing spot of bother’? What light does this attitude reflect on Gerrard?
Answer:
Any other person in Gerrard’s place would have been paralysed with fear when faced with a gun-totting stranger who has entered one’s house to kill him and steal his identity. But Gerrard finds it ‘a amusing spot of bother’ as his cool and unflappable approach makes him handle the situation comfortably and outwit the Intruder with ease.

Question 26.
What is Gerrard’s profession? Quote the parts of the play that support your answer.
Answer:
Gerrard is associated with theatre as a writer, producer and director. He also supplies props and make-up materials to other theatre agencies. The following facts reveal his profession clearly.

• He tells the Intruder that his actions are ‘melodramatic’ but not ‘very original’.
• He welcomes the Intruder as a ‘sympathetic audience’.
• He comments on the Intruder’s ‘inflection of voice’.
• He tells someone over the phone that he cannot deliver the props in time.
• He also tells that person that he had ‘an amusing spot of bother’ which he might put into his next play.

Question 27.
Why was Gerrard’s schedule so irregular?
Answer:
Having a theatrical background, Gerrard devoted time to writing, producing and directing the plays. He also supplied other theatre companies with props and make-up items. Therefore, his schedule was irregular as it had to suit the requirements at the theatres.

Question 28.
Gerrard said, ‘You have been so modest’. Was Gerrard being ironical or truthful?
Answer:
Gerrard’s remark ‘you have been so modest’ was ironical. The Intruder had been boasting of his intelligence and smartness. Hence, Gerrard ironically commented on his modesty and asked him to say something about himself.

Question 29.
What information has the Intruder gathered about Gerrard?
Answer:
The Intruder only knows Gerrard by his last name. He has learnt that, he is the owner of the house in the wilds of Essex. Also, he is a kind of mystery man, who keeps to himself and does not meet anyone. Not many people know about him or visit him.

Question 30.
How did Gerrard fool the Intruder with his false story?
Answer:
Gerrard told the Intruder that he, too, was a criminal on the run. One of his recent crimes had gone wrong and one of his men had been caught. The things which should had been burnt had been found. So, due to that he expected trouble that night. That’s why, he wanted to clear off at the earliest.

Question 31.
How did Gerard lock the Intruder in the cupboard?
Answer:
Having convinced the Intruder that he himself was being wanted by cops and that police could any time break into his home, Gerrard advised the Intruder to escape in his car. When he saw that the Intruder was ready to come with him in his car, Gerrard opened the door of his cupboard and, as the Intruder stepped into the cupboard thinking it was an exit door, Gerrard gave him a push and locked the door from outside.

Question 32.
What precautions did Gerrard take while calling the police?
Answer:
Gerrard was smarter than the Intruder. Having locked the Intruder inside the cupboard, he knocked the revolver out of his hand. To make sure that the Intruder would not break out of the cupboard, Gerrard went to the phone, where he stood with the gun pointed at the cupboard door.

If I Were You Extra Questions and Answers Long Answer Type

Question 1.
Why was Gerrard packing a bag at the beginning of the play? How did it help him to outwit and trap the Intruder?
Answer:
Gerrard was packing a bag in the beginning of the play as he had to deliver some props to some theatrical company for rehearsal. When the Intruder broke into his cottage and threatened to kill him and steal his identity, Gerrard did not lose his cool. He spontaneously concocted a story that he himself was a criminal and was trying to dodge the police. This story was supported by the aura of mystery that surrounded him, his reclusive lifestyle, the bag he had been packing, the disguise outfit, false moustaches etc.

All this misled the Intruder into believing that Gerrard was speaking the truth. He did not doubt Gerrard any longer and unsuspectingly got ready to escape along with him. When Gerrard indicates the door that leads straight to the garage, the Intruder walks into a trap. Hence, the bag played an important role in convincing the Intruder that Gerrard, too, was a criminal like him and was preparing to flee when he broke into his cottage.

Question 2.
Bring out Gerrard’s intelligence, presence of mind and sense of humour. How did these traits help him outwit the Intruder?
Answer:
An intelligent and level-headed person, Gerrard did not show even the slightest of nervousness at the sight of the gun-totting criminal enter his house and threaten to kill him. He knew that his wit and presence of mind would not only help him to manage the crisis but would also contribute towards unnerving the Intruder, and getting the better of him. Keeping the atmosphere light and lively with his sense of humour and funny remarks, Gerrard surprised the Intruder, who had expected him to be afraid.

Once he found out the Intruder was wanted for murder and had been on the run, and thus living in fear, he instantly cooked up a story about his own criminal background. Convincing the Intruder that police would arrive any minute to nab him, he impressed upon the Intruder that they would have to escape immediately. Cleverly, he made him peep into the cupboard saying that it was an escape route.

The moment the Intruder leaned forward to inspect it, Gerrard pushed him into the cupboard and knocked the revolver out of his hand. He closed and locked the door. Thus, his intelligence, sense of humour, and presence of mind turned the tables on the Intruder.

Question 3.
Why did the Intruder find Gerrard’s cooked up story of criminal background convincing?
Answer:
In the beginning, the Intruder suspected every move made by Gerrard. He snubbed him when Gerrard tried to begin a conversation regarding the Intruder’s identity and curtly told him to answer only what was asked. However, he was gullible enough to unsuspiciously walk into Gerrard’s trap because the latter did not lose his cool, and employed his presence of mind to cook up the story that he, too, was wanted by the police.

Actually, Gerrard supported his claim of being a criminal by showing the Intruder his bag, and his disguise outfits and false moustaches etc. The Intruder did not know about the theatrical background of Gerrard and he found his story convincing. He believed that Gerrard was actually trying to evade the police. Hence, the unsuspecting Intruder walked into Gerrard’s trap. This indicates that although he claimed to be the smartest person around, he was in fact not very intelligent. He was outwitted by a smart Gerrard who foiled his plan.

If I Were You Extra Questions and Answers Reference to Context

Read the extracts given below and answer the questions that follow.

Question 1.
Why, this is a surprise, Mr— er—

(a) Who speaks these words and to whom?
Answer:
Vincent Gerrard speaks these words to the Intruder.

(b) Where are they at the time?
Answer:
They are in Gerrard’s cottage, in his sitting room, at the time. The Intruder, who is carrying a revolver has just entered Gerrard’s cottage.

(c) What is the speaker’s tone at the time?
Answer:
The speaker is speaking in a very pleasant tone.

(d) What does this tell you about the speaker?
Answer:
The speaker is a level-headed person. He should have been afraid of the Intruder who was holding a gun, but he was talking in a normal, pleasant manner.

Question 2.
I’m glad you ’re pleased to see me. I don’t think you ’ll be pleased for long. Put those paws up!

(a) Who is speaking these lines and to whom? Where is the conversation taking place?
Answer:
The Intruder is speaking to Gerrard. The conversation is taking place in Gerrard’s lonely cottage situated in the wilds of Essex.

(b) Why is ‘the speaker’ so sure that ‘his listener’ won’t be pleased for long?
Answer:
The speaker is sure that his listener, Gerrard, will not be pleased for long because the speaker plans to kill him and steal his identity.

(c) What does ‘paws’ mean here? Why does the Intruder use the expression?
Answer:
‘Paws’ here stands for ‘hands’. The Intruder wants to convey to Gerrard that he is an American gangster.

(d) Why is the speaker asking the listener ‘to put those paws up’?
Answer:
The Intruder asks Gerrard to put his ‘paws up’ to threaten and intimidate him. He wants to ensure that Gerrard is not able to use his hands for self-defence.

Question 3.
Thanks a lot. You ’ll soon stop being smart. I’ll make you crawl. I want td know a few things, see.

(a) Who is the speaker? Why is he thanking the listener?
Answer:
The Intruder is the speaker here. He is thanking the listener, Gerrard, as the latter had helped him while he was fumbling for a word and Gerrard had suggested the word ‘nonchalant’.

(b) Why does the speaker think that the listener is trying to be smart?
Answer:
The Intruder feels that Gerrard is trying to be smart because instead of displaying any signs of fear, he acts casual and helps the Intruder complete his sentence when the former fails to find the right word.

(c) Why does the speaker expect the listener to soon stop being smart?
Answer:
The speaker feels that Gerrard will be frightened out of his wits the moment he discloses his intention of killing him and will then forget all the witty retorts that he had been making till then.

(d) What does the speaker mean by ‘I’ll make you crawl’?
Answer:
The speaker means that he would bring the listener down on his knees and make him beg for mercy.

Question 4.
At last a sympathetic audience!

(a) Who speaks these words? To whom?
Answer:
Gerrard, the protagonist of the play, speaks these words. He is speaking to the Intruder.

(b) Why does he say it?
Answer:
He wants to throw the Intruder off course by showing him that he does not feel threatened by his presence.

(c) Is he sarcastic or serious?
Answer:
He is certainly sarcastic because he knows that the Intruder wants to gather information about him only to misuse it and he plans to give incorrect information.

(d) Why does the listener wish to know the story of the speaker’s life?
Answer:
The listener is a criminal who resembles Gerrard and wishes to impersonate him. So he wants to know more about him.

Question 5.
I’m sorry. I thought you were telling me, not asking me. A question of inflection; your voice is unfamiliar.

(a) Who is the speaker and who does he speak to?
Answer:
The speaker is Gerrard. He is speaking to the Intruder.

(b) What had the listener asked the speaker?
Answer:
The listener had asked the speaker if he lived in the cottage all by himself.

(c) What does ‘inflection’ mean here? What logic does the speaker give for misinterpreting the inflection of his voice?
Answer:
‘Inflection’ here means ‘tone of voice’. Gerrard says that since the Intruder’s voice was unfamiliar, he couldn’t know whether he was asking a question or telling something.

(d) What do these lines tell us about the speaker?
Answer:
These lines show that the speaker is a very cool-headed man who can think of many ways to elude a question.

Question 6.
That, ’s a lie. You ’re not dealing with a fool. I’m as smart as you and smarter, and I know you run a car. Better be careful, wise guy!

(a) Who is the speaker? Which Tie’ is he talking about?
Answer:
The Intruder is the speaker here. He is talking about the Tie’ that Gerrard told him about not running a car.

(b) Why did the speaker think he was smarter than the listener?
Answer:
The Intruder considered himself smarter because to succeed in his plan of taking on Gerrard’s identity, he had already gathered as much information about Gerrard as he could from the local people.

(c) Why did he warn the listener to be careful?
Answer:
The Intruder wanted to make it clear that Gerrard could not be fool him by telling a lie.

(d) What does the extract reveal about the Intruder?
Answer:
The extract reveals that the Intruder is over-confident about his abilities and that he also under-estimates Gerrard, who is not afraid of him.

Question 7.
You seem to have taken a considerable amount of trouble. Since you know so much about me, won’t you say something about yourself? You have been so modest.

(a) Who speaks these words and to whom?
Answer:
Gerrard speaks these words to the Intruder.

(b) What is his tone when he speaks these words?
Answer:
He is being sarcastic at the time.

(c) Why does he want to know more about the Intruder?
Answer:
He wants to find out more about the Intruder to see if he can get the better of him. He also wants to keep him talking till he receives his telephone call.

(d) What light does this throw on the speaker’s character?
Answer:
The speaker is a quick-thinking person, who does not give way to fear but is looking for a way out of the situation he finds himself in.

Question 8.
I could tell you plenty. You think you ’re smart, but I’m the top of the class round here. I’ve got brains and I use them. That’s how I’ve got where have.

(a) Who speaks these words to whom and in what context?
Answer:
These words are spoken by the Intruder to Gerrard. He utters these words when Gerrard asks him to tell him something about himself

(b) Why does the speaker say “I could tell you plenty”?
Answer:
The Intruder says so because he is over-confident and thinks that he is smart enough to get the better of Gerrard.

(c) What does he mean by ‘the top of the class round here’?
Answer:
The Intruder means to say that no one else is as smart as he is and thatbGerrard, too, is no match for him.

(d) What is his tone at the moment?
Answer:
There is a ring of pride in his words and his ego makes him over-estimate himself and his abilities.

Question 9.
My speciality’s jewel robbery. Your car will do me a treat. It’s certainly a dandy bus.

(a) What does the speaker do? Why does he call it his ‘speciality’?
Answer:
The speaker is a criminal who robs jewellery. He calls it his speciality because robbing jewellery was a pursuit or skill to which he had devoted much time and effort and in which he was an expert.

(b) What does he call ‘a dandy bus’? What does he mean?
Answer:
He calls Gerrard’s car a dandy bus. He means to say that it is a splendid or outstanding car. It will be useful for him and will suit his purpose very well.

(c) What do his words tell you about the speaker?
Answer:
The speaker has made his plans carefully and has found out information about the listener.

(d) What does the speaker intend to do?
Answer:
The speaker uses his brains by planning and committing crimes without getting caught by the police. He now intends to kill Gerrard and assume his identity to escape the law further.

Question 10.
I’m not taking it for fun. I’ve been hunted long enough. I’m wanted for murder already, and they can’t hang me twice.

(a) What ‘step’ is the speaker talking about taking? Why is he taking it?
Answer:
The speaker is talking about taking the ‘step’ of murdering Gerrard. He claims that he is not taking the step for fun but because of his need to escape the police.

(b) By whom has the speaker been hunted? Why?
Answer:
The speaker has been hunted by the police because he is a criminal. He killed a policeman when something went wrong with the job that he did in the town quite a while ago, but since then he is dodging the police.

(c) Why does he say “they can’t hang me twice”?
Answer:
The Intruder has just told Gerrard that he had murdered one man, and that he would not shy away from murdering him too. This is because the police could not hang him twice for two murders.

(d) What light do these lines reflect on the speaker’s state of mind?
The lines reveal that the Intruder does not have any conscience to prick him. He is in a desperate situation now as he fears the punishment he is likely to get if captured.

Question 11.
I’ve got freedom to gain. As for myself I’m a poor hunted rat. As Vincent Charles Gerrard I’m free to go places and do nothing. I can eat well and sleep and without having to be ready to beat it at the sight of a cop.

(a) Why is the speaker a ‘hunted rat’?
Answer:
The Intruder is being chased by the police for having killed a policeman. The fear of being arrested by the police keeps him on the run and he feels that his condition is as miserable as that of a rat being chased.

(b) Why has he chosen to take on Gerrard’s identity?
Answer:
He has chosen to take on Gerrard’s identity because the have a similar height and build and because Gerrard, being a loner, does not meet many people who may catch him out.

(c) Why does the speaker have to run at the sight of a cop?
Answer:
Having killed a cop, the Intruder lives in constant fear of being caught by the police. So, he has to run at the sight of a cop in order to avoid being caught.

(d) What advantage will the speaker have once he impersonates Gerrard?
Answer:
As Gerrard the Intruder will be able to dodge the police. This way he will be able to live in peace and without any fear of the cops.

Question 12.
It brought me to Aylesbury. That’s where I saw you in the car. Two other people saw you and started to talk.
I listened. It looks like you ’re a bit queer — kind of a mystery man.

(a) What is ‘it’? Where did it bring him?
Answer:
‘It’ here refers to the speaker’s dodging the police. While escaping the police he reached Aylesbury.

(b) What did the speaker overhear about the listener? From whom?
Answer:
He overheard two men discuss Gerrard. They referred to him as being strange and a mystery man about whom nothing much was known.

(c) What made the two men conclude that the listener was a mystery man?
Answer:
The two men concluded Gerrard was a mystery man because they did not know much about him. He kept to himself and ordered his supplies on the phone. He did not meet even the tradespeople who delivered the orders. He sometimes went away suddenly and came back just the same.

(d) How did this suit the Intruder’s purpose?
Answer:
This suited the Intruder’s purpose as no one knew Gerrard well enough to recognise him if the Intruder took on his identity. Also, the Intruder would be able to come and go suddenly as Gerrard did.

Question 13.
Don’t be a fool. If you shoot, you ’ll hang for sure. If not as yourself then as Vincent Charles Gerrard.

(a) Why did the speaker say that the listener will be hanged?
Answer:
The speaker said that even if he shot him and took on the speaker’s identity, the listener would be hanged as Gerrard because he was wanted by the police.

(b) What surprise did the speaker give to the listener?
Answer:
Gerrard surprised the Intruder by telling him that he was also a criminal and wanted for murder.

(c) What proof does the speaker give the listener about his being a criminal?
Answer:
Gerrard told the Intruder that he did not meet any trades people and was a bit of a mystery man here today and gone tomorrow because he was a criminal on the run.

(d) What do you think was the speaker’s tone as he spoke to the listener?
Answer:
The speaker’s tone was serious and confidential. The listener was taken in by the speaker.

Question 14.
This is your big surprise. I said you wouldn’t kill me and I was right. Why do you think I am here today and gone tomorrow, never see tradespeople? You say my habits would suit you. You are a crook. Do you think I am a Sunday-school teacher?

(a) What was the big surprise given by the speaker?
Answer:
Gerrard told that the Intruder that he too lived under the threat of being arrested as he too was involved in crime. The Intruder was naturally surprised at this revelation since he was not aware about this aspect of his victim.

(b) What was the speaker right about? Why was he right?
Answer:
Gerrard, the speaker here, was right about the statement that he had made earlier that the Intruder wouldn’t kill him. He was right because the Intruder intended to kill an ordinary person and impersonate him to evade the police. But Gerrard turned out to be a criminal like him. So, killing and impersonating a criminal would not serve the Intruder’s purpose,

(c) Explain the phrase Sunday school teacher? What does the speaker imply by his words?
Answer:
A Sunday school teacher is not just an instructor but is also the responsibility for the spiritual welfare of the students. As such, the Sunday school teacher is an important member of the church and one of high moral standing. By saying he is not a Sunday school teacher, the speaker implies he is a crook.

(d) What light does it throw on the character of the speaker and the listener?
Answer:
The speaker is a quick-thinking cool-headed person, who has retained his presence of mind and lays a trap for the Intruder. He is able to convince the listener. On the other hand, the listener is a gullible person and is taken in by Gerrard’s words.

Question 15.
“I said it with bullets and got away ”.

(a) Who says this?
Answer:
Gerrard, the protagonist of the play ‘If I Were You’, says this.

(b) What does it mean?
Answer:
Gerrard means that he committed a murder with a gun for his escape because things had gone wrong.

(c) Is it the truth? What is the speaker’s reason for saying this?
Answer:
No, it is not the truth. The speaker has concocted a story to befool the Intruder. He shows himself to be a wanted criminal on run from the police so that the Intruder should give up his plan of killing him and taking up his identity.

(d) How was he in imminent danger from the police?
Answer:
One of his men had been caught by the police with some documents.

Question 16.
I have got a man posted on the main road. He’ll ring up if he sees the police, but I don’t want to leave… (telephone bell rings,) Come on! They ’re after us. Through here straight to the garage.

(a) Whose call had Gerrard been expecting?
Answer:
Gerrard had told the person he was speaking to in the beginning to tell someone to call him at once. So, he had been expecting that call.

(b) Whose call had told the Intruder he was expecting?
Answer:
He told the Intruder he was expecting trouble, and had posted a man on the look out who would tell him if the police were coming.

(c) What did he show the Intruder to convince him that he was going to run away?
Answer:
He showed him the packed bag and disguise outfit; false moustaches and what not to show he was ready . to run away.

(d) What is his tone like as he says these words?
Answer:
He says these words in a tone of urgency.

Question 17.
For God’s sake clear that muddled head of yours and let’s go. Come with me in the car. I can use you. If you find it’s a frame, you’ve got me in a car, and you still have your gun.

(a) What does the speaker call the listener’s head “muddled”?
Answer:
The Intruder, who has come to Gerrard’s house to kill him and steal his identity, is told by Gerrard that he, too, is a criminal on the run. The Intruder is thus looking confused.

(b) Where does the speaker invite the other person?
Answer:
The speaker Gerrard is inviting the other person to accompany him in the car and help him escape

(c) What assurance does he give the listener?
Answer:
He tells the Intruder that he has the gun so he can over-power him whenever he feels he has been trapped.

(d) What is in the speaker’s mind?
Answer:
The speaker wants to get the Intruder into a trap where he can hand him over to the police.

In this page, we are providing Online Education for The Living Organisms and Their Surroundings Class 6 Extra Questions and Answers Science Chapter 9 pdf download. NCERT Extra Questions for Class 6 Science Chapter 9 The Living Organisms and Their Surroundings with Answers will help to score more marks in your CBSE Board Exams.

Online Education for Class 6 Science Chapter 9 Extra Questions and Answers The Living Organisms and Their Surroundings

Extra Questions for Class 6 Science Chapter 9 The Living Organisms and Their Surroundings with Answers Solutions

The Living Organisms and Their Surroundings Class 6 Extra Questions Very Short Answer Type

Class 6 Science Chapter 9 Extra Questions And Answers Question 1.
Name the process by which living things obtain energy from food.
Answer:
Respiration

Class 6 Science Chapter 9 Extra Questions Question 2.
Name two aquatic plants which are completely submerged in water.
Answer:
Lily and lotus

Ncert Class 6 Science Chapter 9 Extra Questions Question 3.
Name two terrestrial habitats.
Answer:
Deserts, mountains

The Living Organisms And Their Surroundings Class 6 Questions And Answers Question 4.
What are two components of the habitat?
Answer:
(i) Biotic components
(ii) Abiotic components

Class 6 Science Ch 9 Extra Questions Question 5.
Name one characteristic which tells us that seeds are living things.
Answer:
They grow into a plant when get favourable conditions.

Living Organisms And Their Surroundings Class 6 Extra Questions Question 6.
Name two animals which live in the arboreal habitat.
Answer:
(i) Birds
(ii) Monkey

Ncert Solutions For Class 6 Science Chapter 9 Extra Questions Question 7.
Mention few aquatic plants found in a pond.
Answer:
Water hyacinth, water lily, lotus, Hydrilla, etc.

Class 6 Science Chapter 9 Extra Question Answer Question 8.
Name the habitats of following plants:
(a) Cactus
(b) Hydrilla
Answer:
(a) Cactus: Desert
(b) Hydrilla: Aquatic

Living Organisms And Their Surroundings Extra Questions Question 9.
Name some plants which live in dry areas.
Answer:
Plants like cactus, Acacia, pine, prickly pear live in dry areas.

The Living Organisms Characteristics And Habitats Extra Questions Question 10.
Name two aquatic habitats.
Answer:
Oceans, ponds

Extra Questions For Class 6 Science Chapter 9 Question 11.
Write the names of two sea animal which have nostrils.
Answer:
Dolphins and whales

Living Organisms And Their Surroundings Class 6 Hots Questions Question 12.
What is stimuli?
Answer:
The changes in our surrounding that makes us respond to them are called stimuli.

Chapter 9 Science Class 6 Extra Questions Question 13.
How does gills help a fish?
Answer:
Gills help fish to take-up oxygen dissolved in water.

Living Organisms And Their Surroundings Class 6 Questions And Answers Question 14.
Name an animal in which the exchange of gases, during respiration, take place through its skin.
Answer:
The earthworm breathes through the skin.

Class 6 Chapter 9 Science Extra Questions Question 15.
Name one animal which is adapted to live in water as well as land.
Answer:
Frog

The Living Organisms and Their Surroundings Class 6 Extra Questions Short Answer Type

Question 1.
What is adaptation?
Answer:
The change in specific features or certain habits, which enables a plant or an animal to live in its surroundings is called adaptation.

Question 2.
Define ecology.
Answer:
The study of relationship between living organism and their surroundings is called ecology.

Question 3.
What is environment?
Answer:
Our physical surrounding is called environment.

Question 4.
What is ocean?
Answer:
A very large body of salt water that covers nearly three-fourths of the surface of the earth is called ocean.

Question 5.
Where does a camel lives?
Answer:
A camel lives in a hot desert where there is scarcity of water.

Question 6.
What is habitat?
Answer:
The place or surroundings where a plant or an animal lives is called its habitat.

Question 7.
What is terrestrial habitat?
Answer:
A land based habitat is called a terrestrial habitat.

Question 8.
What is acclimatisation?
Answer:
Small changes that take place in the body of a single organism over short periods, to overcome small problems due to changes in the surroundings, is called acclimatization.

Question 9.
What is reproduction?
Answer:
The process by which living organisms produce new member of their species is called reproduction.

Question 10.
What is excretion?
Answer:
The removal of waste substances from the body of a living organism is called excretion.

Question 11.
What are organisms?
Answer:
The living things are called organisms.

Question 12.
What are biotic components?
Answer:
The living things such as plants, animals and microorganisms in a habitat are its biotic components.

Question 13.
Explain abiotic components.
Answer:
Various non-living things such as rocks, soil, air and water in a habitat constitute its abiotic components.

Question 14.
Why do fish have slippery scales on their bodies?
Answer:
These scales protect the fish from external damage from sharp objects and also help in easy movement through water.

Question 15.
Give three adaptive features in desert plants.
Answer:

1. Well developed root systems.
2. Leaves either very small or converted to spines.
3. Stem is green and fleshy in some plants to carry out photosynthesis.

The Living Organisms and Their Surroundings Class 6 Extra Questions Long Answer Type

Question 1.
List the important characteristics of living things.
Answer:
Following are the important characteristics of living things:

1. Living things need food, air and water.
2. Living things grow.
3. Living things reproduce themselves.
4. Living things respire.
5. Living things are made up of living cells.
6. Living things excrete.
7. Living things have a definite lifespan.

Question 2.
How does the desert animals prevent the loss of water from their body?
Answer:

1. Many animals living in desert remain inactive during the day and become active during night to prevent the loss of water from their body. Such animals are called nocturnal.
2. Certain animals hide themselves in burrows with gelatinous secretion and sleep for about 8 to 9 months in a year. This is called aestivation or summer sleep.

Question 3.
How are plants adapted to aquatic habitat?
Answer:

1. In aquatic plants, roots are much reduced in size and their main function is to hold the plant on its place.
2. The stems may be reduced or long, slender, flexible and spongy. The air spaces in spongy stems make them buoyant.
3. The leaves are thin, large and flat with waxy surface as in lotus and water lily. This protects leaves from wetting and decaying.
4. The leaves of submerged plants are thin and narrow. Hydrilla and Vallisneria are submerged plants and have thin and narrow leaves. Such leaves allow the water to flow without damaging them.

Question 4.
What is respiration?
Answer:
All the living things need energy to grow, move and stay alive. They get this energy from food through respiration. Respiration is the chemical process in which food taken by an organism combines with oxygen to release energy. This energy is used by the organism to carry out its various life processes. Carbon dioxide and water are the other products of the process of respiration.

Question 5.
What is habitat? Explain briefly.
Answer:
All the living organisms (plants and animals) prefer to live in those places which are best suited to their needs. The place or surroundings where a plant or an animal live is called its habitat.

Habitat can be divided into two main groups:
(i) Terrestrial habitat
(ii) Aquatic habitat
A land based habitat is called terrestrial habitat and a water based habitat is called aquatic habitat.

Question 6.
Differentiate between biotic and abiotic components.
Answer:

Abiotic components	Biotic components
1. The term ‘biotic’ means living.	1. The term ‘abiotic’ means non-living.
2. All the living thing in a habitat are biotic components.	2. All non-living things in a habitat are known as abiotic components.
3. Example, plants, animals, etc.	3. Example, air, soil, etc.

Question 7.
Differentiate between xerophytes and hydrophytes.
Answer:

Xerophytes	Hydrophytes
1. Plants are adapted to dry conditions.	1. Plants are adapted to aquatic conditions.
2. Leaves are reduced in size or are modified to spines to reduce transpiration.	2. Leaves are broad for floating or ribbon-shaped to allow water current through it.
3. Roots are long and well developed.	3. Roots are not very well developed. Their main function is just anchoring the plant.
4. Example, Cactus, Acacia	4. Example, Hydrilla, Vallisneria

The Living Organisms and Their Surroundings Class 6 Extra Questions HOTS

Question 1.
What is acclimatization? How it differs from adaptation?
Answer:
Small changes in an organism to overcome small changes in the surrounding is called acclimatization. Whereas adaptation is permanent ability of an organism to make themselves suitable for living in a particular environment. Adaptation is slow and gradual process whereas acclimatization is comparatively a quick method.

Question 2.
Can you name some plant parts, apart from seed, which give rise to a new plant?
Answer:
Leaves, stem, bulb, offset, corm, etc., of some plants can give rise to a new plant.

Question 3.
Is seed a living thing?
Answer:
Yes, seed is a living thing as it has living embryo in it. It is in dormant state which could be break by subjecting it to favourable conditions.

Question 4.
Differentiate between aestivation and hibernation.
Answer:

Aestivation	Hibernation
1. It is summer sleep	1. It is winter sleep.
2. Animals rests in cool or shady and moist place.	2. Animals rest in a warm place.
3. It lasts for hot dry day time as nights are cooler.	3. It is of longer duration and lasts for the whole duration of winter.
4. Activators are snails, earthworms, bees, salamanders, frogs, toads, lizards, snakes, mud turtles, etc.	4. Hibernators are polar bear, kangaroo, mouse, squirrels, etc.

Question 5.
Why do mountaineers carry oxygen cylinders with them?
Answer:
As we go higher the air gets less dense. There is lack of supply of oxygen. Thus, mountaineers carry oxygen cylinders with them.

The Living Organisms and Their Surroundings Class 6 Extra Questions Value-Based (VBQs)

Question 1.
Radha while learning about the characteristics of living organisms studied that living organism excrete their waste out of the body. She remembered that plants are also living organisms. But she wondered how plants excrete their waste. She searched in internet to solve her query.
(a) What are the characteristics of living organisms?
(b) Which component is constituted by plants?
(c) How do plant excrete their waste materials?
(d) What value of Radha is shown here?
Answer:
(a) Living organisms can

1. grow
2. reproduce
3. excrete
4. respire
5. move
6. show response to stimuli

(b) Biotic component
(c) In plants different modes of excretion are present. They loses their gaseous waste through their stomata. Deciduous plants transfer their toxic waste to the leaves which fall on maturity. Excess water is lost through transpiration.
(d) Radha is curious, responsible and hardworking.

Question 2.
Gita lives in Tamil Nadu. Her family planned to visit a hilly area during her school vacation. She was so excited. When she reached there, she experienced breathing problem while walking or running. She got worried that something might have happened to her. But her father consoled her and said, that this is normal and she will be fine after few days of staying there.
(a) Why does Gita experience breathing problem in hilly area?
(b) How will she be fine after some days?
(c) What do this process of adjusting to environment called?
(d) What values of Gita’s father is shown here?
Answer:
(a) In hilly areas oxygen level is very low. So, Gita experienced breathing problem.
(b) Her body will make small changes, like increased red blood cells, to cope with decreased amount of oxygen.
(c) Acclimatisation
(d) Gita’s father is caring, loving, responsible and have scientific aptitude.

Here we are providing Triangles Class 9 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 7 Triangles with Solutions Answers

Triangles Class 9 Extra Questions Very Short Answer Type

Triangles Class 9 Extra Questions Question 1.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
We know that each interior angle of an equilateral triangle is 60°.
∴ Each exterior angle = 180° – 60° = 120°

Class 9 Triangles Extra Questions Question 2.
If in ∆ABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Solution:
Here, ∠A = ∠B + ∠C
And in ∆ABC, by angle sum property, we have
∠A + ∠B + C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the given triangle is a right triangle.

Triangles Extra Questions Class 9 Question 3.
In ∆PQR, PQ = QR and ∠R = 50°, then find the measure of ∠Q.
Solution:
Here, in ∆PQR, PQ = QR
⇒ ∠R = ∠P = 50° (given)
Now, ∠P + ∠Q  + ∠R = 180°
50° + ∠Q + 50° = 180°
⇒ ∠Q = 180° – 50° – 50°
= 80°

Class 9 Maths Chapter 7 Extra Questions With Solutions Question 4.
If ∆SKY ≅ ∆MON by SSS congruence rule, then write three equalities of corresponding angles.
Solution:
Since ∆SKY ≅ ∆MON by SSS congruence rule, then three equalities of corresponding angles
are ∠S = ∠M, ∠K = ∠O and ∠Y = ∠N.

Triangles Class 9 Extra Questions With Solutions Pdf Question 5.
Is ∆ABC possible, if AB = 6 cm, BC = 4 cm and AC = 1.5 cm ?
Solution:
Since 4 + 1.5 = 5.5 ≠ 6
Thus, triangle is not possible.

Triangle Class 9 Extra Questions Question 6.
In ∆MNO, if ∠N = 90°, then write the longest side.
Solution:
We know that, side opposite to the largest angle is longest.
∴ Longest side = MO.

Extra Questions On Triangles Class 9 Question 7.
In ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
Here, in ∆ABC AB = AC ∠C = ∠B [∠s opp. to equal sides of a ∆)
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180° [∵ ∠B = 70°]
⇒ ∠A = 180° – 70° – 70° = 40°

Class 9 Maths Triangles Extra Questions Question 8.
In ∆ABC, if AD is a median, then show that AB + AC > 2AD.
Solution:

Produce AD to E, such that AD = DE.
In ∆ADB and ∆EDC, we have
BD = CD, AD = DE and ∠1 = ∠2
∆ADB ≅ ∆EDC
AB = CE
Now, in ∆AEC, we have
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD [∵ AD = DE]

Triangles Class 9 Extra Questions Short Answer Type 1

Extra Questions On Congruence Of Triangles Class 9 Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:

In ∆DAB and ∆CBA, we have
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Extra Questions Of Triangles Class 9 Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.
Solution:

In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)
In ∆BCD, we have
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Adding (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Extra Questions For Class 9 Maths Triangles Question 3.
In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.
Solution:

In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
AC = AC [common]
∠3 = ∠4 [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since corresponding parts of congruent triangles are equal
∴ BC = CD

Class 9 Maths Chapter 7 Extra Questions Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Triangles Class 9 Important Questions With Solutions Question 5.
In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.
Solution:

Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠s opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
Also, in ∆BDC .
ext. ∠ADB > ∠CBD …(iii)
From (ii) and (iii), we have
∠BDC > CD [∵ sides opp. to greater angle is larger]

Class 9 Maths Ch 7 Extra Questions Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = \(\frac{1}{2}\) AC. Show that ∠ABC is a right angle.
Solution:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = \(\frac{1}{2}\)AC …(i)
Also, BD = \(\frac{1}{2}\)AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°

Triangles Class 9 Extra Questions Short Answer Type 2

Chapter 7 Maths Class 9 Extra Questions Question 1.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:

In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Questions On Triangles For Class 9 Question 2.
In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Ch 7 Maths Class 9 Extra Questions Question 3.
In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC.

Solution:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Triangles Class 9 Extra Questions With Solutions Question 4.
In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.

Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ x = \(\frac{85^{\circ}}{2}\) = 42.59
In ∆ABX, we have x > 30°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (ii)
Hence, from (i) and (ii), we have
CX < AX < BX

Question 5.
In figure, ‘S’ is any point on the side QR of APQR. Prove that PQ + QR + RP > 2PS.

Solution:
In ∆PQS, we have
PQ + QS > PS …(i)
[∵ sum of any two sides of a triangle is greater than the third side]
In ∆PRS, we have
RP + RS > PS …(ii)
Adding (i) and (ii), we have
PQ + (QS + RS) + RP > 2PS
Hence, PQ + QR + RP > 2PS. [∵ QS + RS = QR]

Question 6.
If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.
Solution:
Here, two triangles ABC and BDC having the common
base BC, such that AB = AC and DB = DC.
Now, in ∆ABD and ∆ACD

AB = AC [given]
BD = CD [given]
AD = AD [common]
∴ ΔABD ≅ ΔΑCD [by SSS congruence axiom]
⇒ ∠1 = ∠2 [c.p.c.t.]
Again, in ∆ABE and ∆ACE, we have
AB = AC [given]
∠1 = ∠2 [proved above]
AE = AE [common]
∆ABE = ∆ACE [by SAS congruence axiom]
BE = CE [c.p.c.t.]
and ∠3 = ∠4 [c.p.c.t.]
But ∠3 + ∠4 = 180° [a linear pair]
⇒ ∠3 = ∠4 = 90°
Hence, AD bisects BC at right angles.

Triangles Class 9 Extra Questions Long Answer Type

Question 1.
In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C.

Solution:
Here, AP and DP are angle bisectors of ∠A and ∠D
∴ ∠DAP = \(\frac{1}{2}\)∠DAB and ∠ADP = \(\frac{1}{2}\)∠ADC ……(i)
In ∆APD, ∠APD + ∠DAP + ∠ADP = 180°
⇒ ∠APD + \(\frac{1}{2}\) ∠DAB + \(\frac{1}{2}\)∠ADC = 180°
⇒ ∠APD = 180° – \(\frac{1}{2}\)(∠DAB + ∠ADC)
⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii)
Also, ∠A + ∠B + C + ∠D = 360°
∠B + 2C = 360° – (∠A + ∠D)
∠B + C = 360° – (∠DAB + ∠ADC) ……(iii)
From (ii) and (iii), we obtain
2∠APD = ∠B + ∠C

Question 2.
In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that
(i) DF = BE (i) AM bisects ∠BAD.

Solution:
(i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s]
Also, ∠2 = ∠4
⇒ ∠1 = ∠3
⇒ CE = CF (sides opp. to equals ∠s of a ∆]
∴ DF = BE [∵ BC – CE = CD – CF)

(ii) In ∆ADF and ∆ABE, we have
AD = AB [sides of a square]
DF = BE [proved above]
∠D = ∠B = 90°
⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom]
⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.]
In ∆AMF and ∆AME
AF = AE [proved above]
AM = AM [common]
FM = EM (given)
∴ ∆AMF ≅ ∆AME [by SSS congruence axiom]
∴ ∠7 = ∠8 …(ii) [c.p.c.t.]
Adding (i) and (ii), we have
∠5 + ∠7 = ∠6 + ∠8
∠DAM = ∠BAM
∴ AM bisects ∠BAD.

Question 3.
In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (iv) CM = \(\frac{1}{2}\)AB

Solution:
Given : ∆ACB in which 4C = 90° and M is the mid-point of AB.
To Prove :
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC = 90°
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\)AB
Proof : Consider ∆AMC and ∆BMD,
we have AM = BM [given]
CM = DM [by construction]
∠AMC = ∠BMD [vertically opposite angles]
∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom]
⇒ AC = DB …(i) [by c.p.c.t.]
and ∠1 = ∠2 [by c.p.c.t.]
But ∠1 and ∠2 are alternate angles.
⇒ BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180°
⇒ 90° + CBD = 180°
⇒ ∠CBD = 90°
In ∆DBC and ∆ACB,
we have CB = BC [common]
DB = AC [using (i)]
∠CBD = ∠BCA
∴ ∆DBC ≅ ∆ACB
⇒ DC = AB
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC
⇒ \(\frac{1}{2}\)AB = CM or CM = \(\frac{1}{2}\)AB (∵ CM = \(\frac{1}{2}\)DC)

Question 4.
In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.

Solution:
In ∆BDC, we have ∠DBC = ∠DCB (given).
⇒ CD = BD (sides opp. to equal ∠s of ∆DBC)
Now, in ∆ABD and ∆ACD,
we have AB = AC [given]
BD = CD [proved above]
AD = AD [common]
∴ By using SSS congruence axiom, we obtain
∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [c.p.ç.t.]
Hence, AD bisects ∠BAC of ∆ABC.

Question 5.
Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle.
Solution:

Given : Two As ABC and DEF in which
∠B = ∠E,
∠C = ∠F and BC = EF
To Prove : ∆ABC = ∆DEF
Proof : We have three possibilities
Case I. If AB = DE,
we have AB = DE,
∠B = ∠E and BC = EF.
So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF

Case II. If AB < ED, then take a point Mon ED
such that EM = AB.
Join MF.
Now, in ∆ABC and ∆MEF,
we have
AB = ME, ∠B = ∠E and BC = EF.
So, by SAS congruence axiom,
we have ΔΑΒC ≅ ΔΜEF
⇒ ∠ACB = ∠MFE
But ∠ACB = ∠DFE
∴ ∠MFE = ∠DFE

Which is possible only when FM coincides with B FD i.e., M coincides with D.
Thus, AB = DE
∴ In ∆ABC and ∆DEF, we have
AB = DE,
∠B = ∠E and BC = EF
So, by SAS congruence axiom, we have
∆ABC ≅ ∆DEF
Case III. When AB > ED
Take a point M on ED produced
such that EM = AB.
Join MF
Proceeding as in Case II, we can prove that
∆ABC = ∆DEF
Hence, in all cases, we have
∆ABC = ∆DEF.

Question 6.
In the given figure, side QR is produced to the point S. If the bisectors of ∠PQR and ∠PRS meet at T,
prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Here, QT is angle bisector of ∠PQR

Triangles Class 9 Extra Questions HOTS

Question 1.
Show that the difference of any two sides of a triangle is less than the third side.
Solution:

Consider a triangle ABC
To Prove :
(i) AC – AB < BC
(ii) BC – AC < AB
(iii) BC – AB < AC
Construction : Take a point D on AC
such that AD = AB.
Join BD.
Proof : In ∆ABD, we have ∠3 > ∠1 …(i)
[∵ exterior ∠ is greater than each of interior opposite angle in a ∆]
Similarly, in ∆BCD, we have
∠2 > ∠4 …..(ii) [∵ ext. ∠ is greater then interior opp. angle in a ∆]
In ∆ABD, we have
AD = AB [by construction]
∠1 = ∠2 …(iii) [angles opp. to equal sides are equal in a triangle]
From (i), (ii) and (iii), we have
⇒ ∠3 > ∠4 =
⇒ BC > CD
⇒ CD < BC
AC – AD < BC
AC – AB < BC [∵ AD = AB]
Hence, AC – AB < BC
Similarly, we can prove
BC – AC < AB
and BC – AB < AC

Question 2.
In the figure, O is the interior point of ∆ABC. BO meets AC at D. Show that OB + OC < AB + AC.

Solution:
In ∆ABD, AB + AD > BD …(i)
∵ The sum of any two sides of a triangle is greater than the third side. Also, we have
BD = BO + OD
AB + AD > BO + OD ….(ii)
Similarly, in ∆COD, we have
OD + DC > OC … (iii)
On adding (ii) and (iii), we have
AB + AD + OD + DC > BO + OD + OC
⇒ AB + AD + DC > BO + OC
⇒ AB + AC > OB + OC
or OB + OC < AB + AC
Hence, proved.

Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
A campaign is started by volunteers of mathematical club to boost school and its surrounding under Swachh Bharat Abhiyan. They made their own logo for this campaign. What values are acquired by mathematical club ?
If it is given that ∆ABC ≅ ∆ECD, BC = AE.
Prove that ∆ABC ≅ ∆CEA.

Solution:
Here, it is given that
∆ABC ≅ ∆ECD
AB = CE [c.p.c.t.]
BC = CD [c.p.c.t.]
AC = ED [c.p.c.t.]
Now, in ∆ABC and ∆CEA
BC = AE [given]
AB = EC [proved above]
AC = AC [common]
∴ By using SSS congruence axiom, we have
∆ABC ≅ ∆CEA
Value : Cleanliness and social concerning.

Question 2.
Rajiv, a good student and actively involved in applying knowledge A of mathematics in daily life. He asked his classmate Rahul to make triangle as shown by choosing one of the vertex as common. Rahul tried but not correctly. After sometime Rajiv hinted Rahul about congruency of triangle. Now, Rahul fixed vertex C as common vertex and locate point D, E such that AC = CD and BC = CE. Was the triangle made by Rahul is congruent ? Write the condition satisfying congruence.
What value is depicted by Rajiv’s action?

Solution:
In ∆ABC and ∆DEC, we have
AC = DC [by construction]
BC = EC [by construction]
∠ACB = ∠ECD [vert. opp. ∠s]
By using SAS congruence axiom, we have
∆ABC ≅ ∆DEC
Value : Cooperative learning, use of concept and friendly nature.